What happens to the perimeter of triangle ABC as the two smaller
circles change size and roll around inside the bigger circle?
A 'doodle' is a closed intersecting curve drawn without taking
pencil from paper. Only two lines cross at each intersection or
vertex (never 3), that is the vertex points must be 'double points'
not 'triple points'. Number the vertex points in any order.
Starting at any point on the doodle, trace it until you get back to
where you started. Write down the numbers of the vertices as you
pass through them. So you have a [not necessarily unique] list of
numbers for each doodle. Prove that 1)each vertex number in a list
occurs twice. [easy!] 2)between each pair of vertex numbers in a
list there are an even number of other numbers [hard!]
How many different cubes can be painted with three blue faces and
three red faces? A boy (using blue) and a girl (using red) paint
the faces of a cube in turn so that the six faces are painted in
order 'blue then red then blue then red then blue then red'. Having
finished one cube, they begin to paint the next one. Prove that the
girl can choose the faces she paints so as to make the second cube
the same as the first.
Many thanks to Hannah of The School of St Helen and St Katherine for submitting this solution. Very well done Hannah.
Looking at the diagram of the problem; let the top circle be a, the slightly lower one be b and the bottom one be
c and their centres A, B and C.
Draw a horizontal line to the left of C and extend it to the edge of the square.
Now draw a vertical line down from A and call the point where it meets the line you have just drawn D so that angle ADC = $ 90^o $
It is also worth noting that triangle ABC is equilateral and the whole box has a line of symmetry running from the top left hand corner of the square to the bottom right.
The angle DAE (on diagram) is $ 90^o $
and the angle CAB is $ 60^o $ because it is an equilateral triangle. Therefore angle DAC + angle EAB equals $ 30^o; (90^o - 60^o) $ meaning that angle DAC is $ 15^o $; as it equals EAB.
Now we can use trigonometry to work out the length AD (from A, vertically down to D). Using the triangle ADC, the hypotenuse of which is 1cm (two radii), ($ 90^o \times 1 = 0.965...$)
From C to the base of the square is 0.5cm (radius), which is equal to the distance from D to the base of the square. From A to the top of the square is also 0.5cm (radius)and you can add these two values on to AD to get a side of the square.
Therefore one side of the square $= 0.5+0.5 (a+c$ radii$)+0.965... =1.966$ cm(3d.p.)